Strongly Maximal Submodules with A Study of Their Influence on Types of Modules

Let S be a commutative ring with identity, and A is an S-module. This paper introduced an important concept, namely strongly maximal submodule. Some properties and many results were proved as well as the behavior of that concept with its localization was studied and shown.


Definition (2.6)
Let B be a submodule of an S-module A , the closure of B is denoted by CL(B) = { x ∈ A : [B:(x)] essential in S }. It is clear that CL(B) is a submodule of A containing B. That is B ⊆ CL(B) [7] .

S3: Strongly Maximal Submodules with its advantages
In this section, the concept of strongly maximal submodule (for short, SM-submodule) was introduced, which was a generalization of the concept the strongly maximal ideal in a ring S. Several examples and properties were proved also a lot of characterizations, and different results were presented.
Let us start with our basic definition.

Definition (3.1)
Let A be an S-module, and B be a non-zero proper submodule of A. Then B is named strongly maximal submodule (for short SM-submodule) if and only if, for every non-zero ideal E of S implies A/E 2 B is a regular module.

Examples and Remarks (3.2)
1. All the following modules have no SM-submodules.
(i) Z as a Z-module.
(ii) Zp as a Z-module, p is a prime number.
(iii) Zp as a Zp-module, p is a prime number.
2. Every simple S-module has no SM-submodule. But the opposite is not true and the following example shows that: The module A=Z4⊕Z as a Z-module. Since A has no SMsubmodules , A is not a simple module. Also notice examples (ii) and (iii) in no.(1) .
3. It is important to note that it is not necessary that all modules contain SM-submodules; for example Zp ∞ as-Z-module.
Since, all the submodules of Zp ∞ are of the form <1/p i + Z>, where p is a prime number and i= 0, 1, 2… Now, we write N = <1/p i +Z> and let E be an ideal of Z. If we take E= <1>, then, In general, all submodules of Z6 as a Z6-module are SM-submodules. 5. In Z10 as a Z20-module, the submodule B = <5> is an SM-submodule. Since if we take E = <1> , <2> , <10> , <4> , <5> ,where E is ideal of Z20 , then Z20/E 2 <5> is a regular module for all ideal E of Z20 . This ends the proof of example. 6. Consider Z4 as a Z-module. The submodule <2> is not SM-submodule of Z4. Since Z4/E 2 <2> is not regular, E is an ideal of Z. To prove that, take E = <4>. Then, Z4/<4> 2 <2> ≅ Z4 is not regular module.  10. Consider A = Z6⊕Z as a Z-module. Then the submodule B = <3>⊕<2> of A is not SM- 11. Every SM-submodule is maximal but the opposite is not true and the following example shows that: The submodule <2> of a Z-module Z4 is a maximal submodule in Z4 but it is not a SM-submodule, see no.(6) .  (11), we obtain that every SM-submodule of an Smodule A is a semimaximal while the converse is not true in general and the following shows that: Let 6Z be a submodule of a Z-module Z . Then, 6Z is a semimaximal submodule of Z. Since 6Z =2Z∩5Z where 2Z, 5Z are maximal submodules of Z. But 6Z is not an SMsubmodule of Z. Since Z/(2Z) 2 (6Z) = Z/24Z ≅ Z24 is not a regular module .

Proposition (3.3)
Let B, D be two submodules of an S-module A with B⊆D. Then B is SM-submodule in D when B is SM-submodule in A.

Proof:
Suppose B is SM-submodule in A, then A/E 2 B is regular module for every non-zero ideal E of S. Since D/E 2 B is a submodule of A/E 2 B (Notice, E 2 B ⊆ B ⊆ D), and hence by proposition (2.1), D/E 2 B is a regular submodule of A/E 2 B. Thus B is SM-submodule in D.
Next, we will give an application to a proposition (3.3)

It clear that B⊆ [B A
A : ] ⊆ A. Therefore, by using proposition (3.3), we conclude that B is SM-submodule in A. Now, we will give the sufficient condition for a submodule to not be SM-submodule.

Proposition (3.5)
Let A be an S-module. If A is cyclic module (if A=Sx for some x∈A) and anns(x) is maximal ideal of S, then A has no SM-submodule.

Proof:
Since A= Sx for some x∈A, then A is isomorphic to a factor module of S by proposition (2.2). We can define f : S → A such that f(r) = rx . It is easily to show that f is welldefine and epimorphisim, by the first fundamental theorem of isomorphism S/Ker f ≅ A. Next, Ker f ={r∈S : f(r) =0A } = {r∈S : rx=0A } =anns(x) That is S/anns(x) ≅A. Also, we have anns(x) is maximal ideal of S which implies S/anns(x) is simple and hence A is simple. Therefore, by examples and remarks ((3.2) No. (2)), we get the result.
Next is the application of proposition (3.5)

Corollary (3.6)
If a non-zero prime and semi-simple S-module A, then A has no SM-submodule.

Proof:
Suppose that A is a prime and semi-simple module, then, we obtain A is simple module. To prove this, assume that A is not simple which implies A is a direct sum of simple Smodules. Then, there exists a simple module M1 and M2 which are a direct summand of A with M1 ≠ M2. M1≅ S/E, M2≅ S/D where E and D are maximal ideals of S, by proposition (2.3). But A is prime module, then anns(M1) = E= anns(A) and anns(M2) = D =anns(A).Thus E = D which implies that M1 = M2, and this is a contradiction. Hence, A is a simple module and by using proposition (3.5), we have A has no SM-submodule.
As a direct of corollary (3.6), we have the following.

Corollary (3.7)
Ibn Al-Haitham Jour. for Pure & Appl. Sci. 5 3 (1)2022 89 Let P be a prime and semimaximal submodule of an S-module A. Then the quotient module by P has no SM-submodule.

Proof:
Since P is a semimaximal submodule of A, then by definition (2.5), we have A/P is a semisimple S-module. On the other hand, P is a prime submodule of A, then, by [3, proposition (1.1.51)] we obtain that A/P is a prime S-module, and hence by corollary (3.6), then A/P is a simple S-module and hence A/P has no SM-submodule.

S4: The behavior of SM-submodules under localization.
Let K be a subset of a ring S, W is multiplication closed if the two condition hold: 1. I ∈ W . 2. xy ∈ W for every x , y ∈ W . We know that every proper ideal E in S is prime if and only if S-E is multiplicatively closed, see [8].If A is an S-module and W be a multiplicatively closed on S such that W≠<0> , then Sw be the set for all fractional r/w where r ∈S and w ∈W and Aw be the set of all fractional m/w where m ∈A and w ∈W . For m1, m2∈ A and w1,w2∈ W , m1/w1=m2/w2 if and only if ∃ t ∈ W such that t(w1m1-w2m2)= 0. Also, we can make Aw in to Sw-module by setting m1/w1+m2/w2 =(w2m1+w1m2)/w1w2 and (r/w1) (m1/w2) = rm1/w1w2 for every m1,m2 ∈A and every r ∈ S , w1,w2 ∈ W. If W =S-E where E is a prime ideal, we used AE instead of Aw and SE instead of Sw. If a ring has only one maximal ideal, then it is called a local ring. Hence SE is often called the localization of S at E, similar AE is the r/1,Ɐ r ∈ S and :A →Aw such that (m)=m/1 ,Ɐm∈A. Furthermore, if B is a submodule of an S-module A and W be a multiplicatively closed in S, then Bw = {n/w: n ∈B, w∈W} be a submodule on Swmodule , see [8].
In this section we study the behavior of an SM-submodule under localization and several of results have been proved.
The following lemma is needed in our next result.

Lemma (4.1) [10]
Let A be an S-module and B, L are two submodules of A. Then, B=L if and only if BP=LP for every maximal ideal P of S.
The following proposition study the relationship between a module A and its locally and prove that they are equivalent.

Proposition (4.2):
Let A be an S-module and B is nonzero proper submodule of A. Then, BP is SMsubmodule of an SP-submodule AP if and only if B is SM-submodule of an S-module A.

Proof:
Suppose that B is nonzero proper submodule of A. We must prove that A/E 2 B is a regular S-module for every nonzero ideal E of S; that is, every submodule of A/E 2 B is pure. Let L/E 2 B be a submodule of A/E 2 B. It is clear that I(L/E 2 B) ⊇ I(A/E 2 B)⋂(L/E 2 B) where I is an ideal of S. Now, to prove I(L/E 2 B) ⊆ I(A/E 2 B)⋂(L/E 2 B). Let x ∈ I(L/E 2 B). Then x = ∑ a n i=1 i(li+E 2 B) . Therefore xs/s = (∑ a .

Proposition (4.3):
Let L, B be two finitely generated submodules of an S-module A. If LP, BP are SMsubmodules of AP, then L⋂B is an SM-submodule of A.

Proof:
Since L, B are two finitely generated submodules of A, then by {10,p24}, [LP:BP] + [BP:LP] = SP for every maximal ideals P of S. Thus, LP⋂BP = LP or LP⋂BP = BP, but LP and BP are SM-submodules, then LP⋂BP is an SM-submodule, and we have LP⋂BP = (L⋂B)P. Therefore (L⋂B)P is an SM-submodule and by proposition(4.2), L⋂B is an SM-submodule of A.

Proposition (4.4):
Let L, B be two finitely generated submodules of an S-module A. Then, L+B is an SMsubmodules of A, if LP ,BP are SM-submodules of an SP-module AP .

Proof:
Let L, B be two finitely generated submodules of A. Then by {10, p24}, we have [LP:BP]+[BP:LP] = SP for every maximal ideal P of S. Let y1 ∈ [LP:BP] and y2 ∈ [BP:LP] such that y1+y2 =1 = unity of SP. Then, either y1 is a unit element or y2 is a unit element (Since SP is local ring). Therefore [LP:BP] = SP or [BP:LP] = SP and hence either LP ⊆ BP or BP ⊆ LP which implies LP + BP = LP or LP + BP = BP ,but LP , BP are SM-submodules of AP . Thus LP + BP is an SM-submodule and (L+B)P is an SM-submodule and by proposition(4.2) , L+B is an SM-submodule of A.

5.Conclusion
The conclusion of this work is to study an important concept, namely strongly maximal submodule. Some properties and many results were proved and the behavior of that concept with its localization were studied and shown.